# Half Area

The horizontal red line divides this equilateral triangle into two shapes of equal area. How long is the red line?

## Problem

This equilateral triangle has sides of length 2 cm.

Image

The red line, which is parallel to the base, divides this equilateral triangle into two shapes of equal area.

How long is the red line?

## Student Solutions

**Using the relationship between area and length scale factors**

The area scale factor between the smaller triangle and the larger triangle is $2$.

The area scale factor is always the square of the length scale factor, so the length scale factor must be $\sqrt{2}$.

So the red length must be $2\div \sqrt{2}=\sqrt{2}$ cm.

**Finding the areas of the pieces**

Let the red length be $x$ cm.

The perpendicular heights of the equilateral triangles can be found using Pythagoras' theorem.

Image

For the larger triangle, half of the base is 1 cm, and the hypotenuse is 2 cm, so, calling the height $h$ cm, $1^2+h^2=2^2$, so $h^2=3$, so $h=\sqrt{3}.$

Using the scale factor $\dfrac{x}{2}$ between the two triangles, or applying Pythagoras' theorem again to the smaller triangle, the height of the smaller triangle must be $\dfrac{x\sqrt{3}}{2}.$

So the area of the larger triangle is $\dfrac{1}{2}\times 2\times\sqrt{3}$ cm$^2$ and the area of the smaller triangle is $\dfrac{1}{2}\times x\times \dfrac{x\sqrt{3}}{2}$ cm$^2.$

So, since the area of the smaller triangle is half of the area of the larger triangle, $$\begin{align}\frac{1}{2}\times 2\times\sqrt{3}&=x\times \dfrac{x\sqrt{3}}{2}\\

\Rightarrow 1&=\dfrac{x^2}{2}\\

\Rightarrow x^2&=2\\

\Rightarrow x&=\sqrt{2}\end{align}$$

So the red length must be $2\div \sqrt{2}=\sqrt{2}$ cm.